poj3261之可重叠的k次最长重复子串

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8511		Accepted: 3869
Case Time Limit: 2000MS

Description

Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.

Input

Line 1: Two space-separated integers:  N and  K 
Lines 2.. N+1:  N integers, one per line, the quality of the milk on day  i appears on the  ith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at least  K times

Sample Input

8 2
1
2
3
2
3
2
3
1

Sample Output

4

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=20000+10;
int *rank,sa[MAX],height[MAX],temp[MAX];
int wa[MAX],wb[MAX],wm[MAX];

struct Node{
	int x,id;
	bool operator<(const Node &a)const{
		if(x == a.x)return id<a.id;
		return x<a.x;
	}
}r[MAX];

bool cmp(int *r,int a,int b,int l){
	return r[a] == r[b] && r[a+l] == r[b+l];
}

void makesa(Node *r,int *sa,int n,int m){
	int *x=wa,*y=wb,*t,i,j,p;
	sort(r,r+n);//由于这里最初的m很大所以不采取基数排序,而是采用快速排序,当然也可以先离散化再基数排序 
	for(i=0;i<n;++i)sa[i]=r[i].id;
	for(i=1,x[sa[0]]=0;i<n;++i)x[sa[i]]=(r[i].x == r[i-1].x?m-1:m++);
	/*for(i=0;i<m;++i)wm[i]=0;
	for(i=0;i<n;++i)wm[x[i]=r[i]]++;
	for(i=1;i<m;++i)wm[i]+=wm[i-1];
	for(i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;*/
	for(i=0,j=1,p=0;p<n;j=j*2,m=p){//进行基数排序,当然也可以用快速排序,但是这里基数排序更快 
		for(p=0,i=n-j;i<n;++i)y[p++]=i;
		for(i=0;i<n;++i)if(sa[i]>=j)y[p++]=sa[i]-j;
		for(i=0;i<m;++i)wm[i]=0;
		for(i=0;i<n;++i)wm[x[y[i]]]++;
		for(i=1;i<m;++i)wm[i]+=wm[i-1];
		for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i];
		for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){
			x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
		}
	} 
	rank=x;
}

void calheight(int *r,int *sa,int n){
	for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k){
		for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k);
	}
}

int main(){
	int n,k;
	while(cin>>n>>k){
		for(int i=0;i<n;++i)cin>>temp[i],r[i].x=temp[i]+1,r[i].id=i;
		r[n].x=0,r[n].id=n;
		makesa(r,sa,n+1,1);
		calheight(temp,sa,n);
		int l=1,r=n,mid,sum=0,num=1;
		while(l<=r){
			mid=l+r>>1;
			num=1;
			for(int i=1;i<=n;++i){
				if(height[i]>=mid)++num;
				else num=1;
				if(num>=k)break;
			}
			if(num>=k)l=mid+1;
			else r=mid-1;
		}
		cout<<r<<endl;
	}
	return 0;
}